Lines of Invariant Points, Invariant Lines

Invariant means something that doesn't change. A Linear Transformation will transform all the points on a grid, but invariant points will remain in their original position. We will look at invariant points, lines of invariant points, and invariant lines under linear transformations.

Prerequisites

You'll need to have an understanding of Linear Transformations.

Invariant Points

Under the transformation T, point A(2,3) is invariant. The image of A(2,3) is A*(2,3), with the same coordinates.

Say we have a linear transformation represented by matrix $M$. Then, an invariant point is any point which stays the same under the transformation.

In other words, the point A with position vector ${\underline {x}}$ is invariant if $M{\underline {x}}={\underline {x}}\quad (*)$.

Since position vector ${\underline {x}}$ is really shorthand (in 2-dimensions) for ${\begin{pmatrix}x\\y\end{pmatrix}}$, it can be clearer to write this as $M{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}x\\y\end{pmatrix}}$.

In 3-dimensions, no problem: we could say $M{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$.

For example, let's consider the matrix $M={\begin{pmatrix}2&4\\1&5\end{pmatrix}}$ and the point $B(4,-1)$.

We observe that ${\begin{pmatrix}2&4\\1&5\end{pmatrix}}{\begin{pmatrix}4\\-1\end{pmatrix}}={\begin{pmatrix}4\\-1\end{pmatrix}}$.

Therefore, B(4,-1) is an invariant point (under the linear transformation M). The image of B is B itself.

Note that the origin (0,0) is always an invariant point - this is a property of linear transformations.

Lines of invariant points

A Line of Invariant Points - A(2,3), O(0,0), and B(-1,${\frac {-3}{2}}$) are invariant, lying on the line $3x=2y$. Points away from the line, such as C(2,-3), don't have to be invariant.

A line of invariant points is a straight line where all the points are invariant.

The word 'invariant' applies to the individual points here - which happen to form a line.

With few exceptions, a line of invariant points must normally pass through the origin, so we can give it the (2 dimensional) equation $y=mx$.

Warning: I would recommend eigenvalues and eigenvectors as an alternative method here.

By taking the equation $(*)$ above, and substituting $y=mx$, we can say: $M{\begin{pmatrix}x\\mx\end{pmatrix}}={\begin{pmatrix}x\\mx\end{pmatrix}}$

Invariant Lines

An invariant line. Notice that none of the points themselves are invariant, apart from the origin - the others have been stretched away from the origin. Since lines are continuous and infinitely long in both directions, there will always be points to 'fill the gaps' in the image of the line. The image line will remain continuous and infinitely long.

When we say invariant line, we mean that the line stays the same. But, the points on the line could move along the line. The individual points can change - there don't have to be invariant points. This is going to look like stretching out the points on the line, away from (or towards) the origin.

The point $(x_{0},y_{0})$, on the line, will move to $(x_{1},y_{1})$, also on the line.

Since we are talking about the line $y=mx$, we can say $(x_{0},mx_{0})$ will transform to $(x_{1},mx_{1})$.

Warning: I would recommend eigenvalues and eigenvectors as an alternative method here, which will also be easier to deal with in 3D.

This gives us a method for finding invariant lines. Solve $M{\begin{pmatrix}x_{0}\\mx_{0}\end{pmatrix}}={\begin{pmatrix}x_{1}\\mx_{1}\end{pmatrix}}$.

Example Question

Example: Find the two invariant lines in the linear transformation represented by T = ${\begin{pmatrix}3&2\\2&3\end{pmatrix}}$.

Note: You should consider finding the eigenvectors as an alternative method here.

Since this is a linear transformation with a finite number of invariant lines, let's go ahead and assume that relevant lines include the origin. So, any such lines will be of the form $y=mx$, which means we can avoid working with $+c$ as an unknown. There are exceptions to this, but it makes sense to first identify lines which pass through the origin.

Now, for any point, $T{\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}$.

Therefore, for a point on an invariant line, ${\begin{pmatrix}3&2\\2&3\end{pmatrix}}{\begin{pmatrix}x_{0}\\mx_{0}\end{pmatrix}}={\begin{pmatrix}x_{1}\\mx_{1}\end{pmatrix}}$

Multiplying out the matrices, we get two equations:

${\begin{aligned}3x_{0}+2mx_{0}&=&x_{1}\qquad &(1)\\2x_{0}+3mx_{0}&=&mx_{1}\qquad &(2)\end{aligned}}$

And substituting $x_{1}$ from $(1)$ into $(2)$:

${\begin{aligned}&2x_{0}+3mx_{0}&=&m(3x_{0}+2mx_{0})\qquad &\\\implies &2x_{0}+3mx_{0}&=&3mx_{0}+2m^{2}x_{0}\qquad &\\\end{aligned}}$

Notice that all the terms now contain $x_{0}$. Since the case when $x_{0}=0$ will represent the origin, which we are not looking for here, and is always invariant anyway, we can divide both sides by $x_{0}$:

${\begin{aligned}\implies {}&2+3m&=&3m+2m^{2}&\\\implies {}&2m^{2}&=&2&\\\implies {}&m^{2}&=&1&\\\end{aligned}}$

$\implies {}m=1{\text{ or }}m=-1$

Therefore, our invariant lines are $y=x$ and $y=-x$.